【Leetcode】Largest Rectangle in Histogram (Water)

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.



Above is a histogram where width of each bar is 1, given height = 

[2,1,5,6,2,3]

.



The largest rectangle is shown in the shaded area, which has area = 

10

 unit.

For example,

Given height = 

[2,1,5,6,2,3]

,

return 

10

.
解题思路

这道题一开始的思路是从一个点开始向两边同时扩大，扩到不能再扩了就结束
什么时候是不能再扩了呢？

就是这个点的左右两边都比它本身小的时候

比如图上1这个点就能像两边无限扩大，5就只能向右扩大到5和6.

到底如何求这个面积呢，这里提出三个关键点

1. 栈里面的元素必须按递增排列

2. 定义一个i，它的作用是用来求宽度的

3. 根据1，2一定能求出全面积

为了保证i处的位置一定要是最小，我们在末尾加上一个0，这样i结束的时候一定是最小。

public static int largestRectangleArea(int[] height) {
if (height == null || height.length == 0)
return 0;

ArrayList<Integer> newHeight = new ArrayList<Integer>();
for (int i = 0; i < height.length; i++)
newHeight.add(height[i]);
newHeight.add(0);

LinkedList<Integer> stack = new LinkedList<Integer>();
int maxArea = 0;
int i = 0;

while (i < newHeight.size()) {
if (stack.isEmpty()
|| newHeight.get(stack.peek()) < newHeight.get(i))
stack.push(i++);
else {
int top = stack.pop();
maxArea = Math.max(maxArea,
newHeight.get(top) * (stack.isEmpty() ? i : i - stack.peek() - 1));
}
}
return maxArea;
}

面积公式由height[top]和i或者i~peek的距离（不包含i和peek）决定。

maxAera=newHeight.get(top) * (stack.isEmpty() ? i : i - stack.peek() - 1)